Data Structures...7/12/2008 6 Graphs • In an undirected graph, G, two vertices v1 and v2 are said...

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Data Structures

Elementary Graph Algorithms

BFS, DFS & Topological Sort

Tzachi (Isaac) Rosen

The 7 Bridges of Konigsberg


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• A graph, G = (V, E), consists of two sets:

Graphs

– V is a finite non‐empty set of vertices.

– E is a set of pairs of vertices, called edges.


Graphs

• In an undirected graph Th i f i d d i– The pair of vertices are unordered pairs.

– Thus, the pairs (v1, v2) and (v2, v1) are the same.

– No reflective (self) edges.

• In a directed graph – The edges are represented by a directed pair (v1, v2).

– Therefore (v2, v1) and (v1, v2) are two different edges

– v1 is the tail and v2 the head of the edge.


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Graphs

• G1 and G2 are undirected.• G3 is a directed graph.

G ({1 2 3 4} {(1 2) (1 3) (1 4) (2 3) (2 4) (3 4)} )• G1 = ({1,2,3,4}, {(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)} )• G2 = ({1,2,3,4,5,6,7}, {(1,2),(1,3),(2,4),(2,5),(3,6),(3,7)} )• G3 = ({1,2,3}, {<1,2>, <2,1>, <2,3>})


Graphs

• The maximum number edges in– An undirected graph with n vertices is n(n ‐ 1)/2

– A directed graph with n vertices is n2

• An n vertex undirected graph with exactly n(n‐1)/2 edges is said to be complete


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Graphs• If (v1,v2) is an edge of a graph G, then

– We shall say the vertices v1 and v2 are adjacentAnd that the edge (v v ) is incident on vertices v and v– And that the edge (v1,v2) is incident on vertices v1 and v2

• If (v1,v2) is a directed edge, then vertex v1 will be said to be adjacent to v2, while v2 is adjacent from v1.

• The degree of a vertex is– The number of edges incident to that vertex

• In case G is a directed graph, we define– the in‐degree of a vertex v to be the number of edges for which v is the head.– The out‐degree is defined to be the number of edges for which v is the tail.

•


Graphs

• A subgraph of a graph G is a graphgraph G is a graph G' = (V’, E’) such that– V' V

– G' G.


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Graphs

• A path from vertex vp to vertex vq (vp vq) in graph G is – A sequence of vertices vp,vi1,vi2, ...,vin,vqp i1 i2 in q

such that– (vp,vi1),(vi1,vi2), ...,(vin,vq) are edges

• The length of a path is the number of edges on it• A simple path is a path in which

– All vertices except possibly the first and last are distinct


Graphs

• A cycle is a path in which– The first and last vertices are the same

• A simple cycle – same with simple path

• When the graph is directed, we add the prefix "directed" to the terms


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Graphs

• In an undirected graph, G, two vertices v1 and v2 are said to be connected

If th i th i G f t– If there is a path in G from v1 to v2.– Since G is undirected, this means there must also be a path from v2 to

v1. • An undirected graph is said to be connected

– If for every pair of distinct vertices vi, vi in there is a path from vi to vj in G

• A (connected) component of an undirected graph is a maximal connected subgraph

• A tree is a connected acyclic undirected graph


Graphs


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Graph Representation

• Two common ways to represent a graph ( ith di t d di t d)(either directed or undirected):– Adjacency lists.

– Adjacency matrix.


Adjacency Lists

• Array Adj of |V| lists,– One per vertex.– Vertex u’s list has all vertices v such that (u, v) E.– Works for both directed and undirected graphs.

• Space: Θ(V + E)• Space: Θ(V + E).– When expressing the running time, we’ll drop the cardinality.

• Time to list all vertices adjacent to u: Θ(degree(u)).• Time to determine if (u, v) E: O(degree(u)).


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Adjacency Matrix

•

• Space: Θ(V2).• Time to list all vertices adjacent to u: Θ(V).• Time to determine if (u, v) E: O(1).


Breadth‐First Search

• Given a graph G = (V, e) and a vertex s E– Which vertex is reachable from s

– What is the shortest distance to it

– What is the shortest path


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• Algorithm Idea:– Send a wave out from s:Send a wave out from s:

• First hits all vertices 1 edge from s.• From there, hits all vertices 2 edges from s.• Etc.

– Use FIFO queue Q to maintain wave front.• v Q if and only if wave has hit v but has not come out of v yet.



• Input:– Graph G = (V, E), either directed or undirected.– Source vertex s V.

• Output: – d[v] = distance to all v V from s.

• If v V is not reachable from s, d[v] = ∞.– π[v] = u such that (u, v) is last edge on shortest path s v.

• If v V is not reachable from s, π[v] will be null.• The set of edges {(π[v], v) : v ≠ s} forms a tree, such that u is v’s predecessor.

• Auxiliary Means:– every vertex has a color:

• White ‐ undiscovered• Gray ‐ discovered, but not finished (not done exploring from it)• Black ‐ finished (have found everything reachable from it)


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BFS(G,s)

for (each vertex u V[G] {s}) dofor (each vertex u V[G] ‐ {s}) docolor[u] = white, d[u] = ∞, π[u] = null

Q = {s}, color[s] = gray, d[s] = 0 , π[s] = null

while (Q ≠ ) dou = dequeue(Q)

for (each v Adj[u] ) do( j[ ] )if (color[v] = white) then

enqueue(Q, v), color[v] = gray, d[v] = d[u] + 1, π[v] = u

color[u] = black


Complexity: O(V + E)

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• The shortest‐path distance from s to v, δ(s, v), is:The minimum length of the paths from s to v– The minimum length of the paths from s to v,or else

– ∞, if there is no path from s to v.

• A path of length δ(s, v) from s to v is said to be a shortest path from s to v.



• Theorem:– Suppose that BFS is run on graph G = (V, E) from a source s V,Suppose that BFS is run on graph G (V, E) from a source s V,

then

– It discovers every vertex v V that is reachable from s.

– Upon termination, d[v] = δ(s, v) for all v V.– For any vertex v ≠ s that is reachable from s,

• One of the shortest paths from s to v is a shortest path from s to π[v] followed by the edge (π[v], v).

H h h ( [ [ ]]) ( [ [ ]] [ ]) ( [ ] ) i f h• Hence, the path (s, π[…π[v]]), …, (π[π[v]], π[v]), (π[v], v) is one of the shortest path from s to v.


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Breadth‐First Search• Proof:

– We will use the following two facts:• Fact 1: Upon termination for each v V d[v] ≥ δ(s v)• Fact 1: Upon termination, for each v V, d[v] ≥ δ(s, v).• Fact 2: If vertex vi is enqueued before vj during the execution then d[vi] ≤ d[vj].

– Assume, for the purpose of contradiction, that there is v V such that upon termination d[v] ≠ δ(s, v).

– If there are more then one, take v with minimum δ(s, v).– Clearly v ≠ s.– By fact 1, d[v] ≥ δ(s, v), and thus, from the assumption, d[v] > δ(s, v).– Vertex v must be reachable from s, for if it is not, then δ(s, v) = ∞ ≥ d[v].– Let u be the vertex immediately preceding v on a shortest path from s to v, so

that δ(s v) = δ(s u) + 1that δ(s, v) = δ(s, u) + 1.

– Since δ(s, u) < δ(s, v), and because of how we chose v, we have d[u] = δ(s, u).– Putting these together, we have d(v) > δ(s, v) = δ(s, u) + 1 = d[u] + 1.


Breadth‐First Search• Consider the time when vertex u dequeue from Q.• At this time, vertex v is either white, gray, or black.

If i hit th th BFS t d[ ] d[ ] 1 t di ti– If v is white, then the BFS sets d[v] = d[u] + 1, contradiction.– If v is black, then it was already removed from the queue and, by fact 2, we

have d[v] ≤ d[u], contradiction.– If v is gray, then it was painted gray upon dequeuing some vertex w, which was

removed from Q earlier than u and for which d[v] = d[w] + 1.– By fact2, however, d[w] ≤ d[u], and so we have d[v] ≤ d[u] + 1, contradiction.

• Thus we conclude that d[v] = δ(s, v) for all v V.• All vertices reachable from s must be discovered, for if they were not, they

would have infinite d values.• if π[v] = u, then d[v] = d[u] + 1. Thus, we can obtain a shortest path from s

to v by taking a shortest path from s to π[v] and then traversing the edge (π[v], v).

• By induction, the path (s, π[…π[v]]), …, (π[π[v]], π[v]), (π[v], v) is one of the shortest path from s to v.


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Printing The Shortest Path

PrintPath (G, s, v)if ( ) thif (v = s) thenprint s

else if (π[v] = null) thenprint "no path from" s "to" v "exists"

elseelseprintPath (G, s, π[v])print v


Depth‐First Search

• Given a graph G = (V, E)– Draw G as a forest of sub graphs

– Say which vertex is a descendant of another in the forest.

– Detect cycles

– Topological Sortp g


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• Algorithm Idea:– Search deeper in the graph whenever possible:

• Start from an arbitrary unvisited vertex.

• Explore out one of the undiscovered edge of the most recently discovered vertex v.

• When all of v's edges have been explored, backtracks, and continueand continue.

• Start all over again.



• Input:– Graph G = (V, E), either directed or undirected.

• Output: – d[v] = discovery time & f [v] = finishing time.

• A unique integer from 1 to 2|V| such that 1 ≤ d[v] < f [v] ≤ 2|V|.– π[v] = u such that u is the predecessor to v in the visit order.

• The predecessor subgraph Gπ = (V, Eπ), where Eπ={(π[v],v) : π[v]≠null} forms a forest composed of several trees.

• Auxiliary Means:– every vertex has a color:

• White ‐ undiscovered• Gray ‐ discovered, but not finished (not done exploring from it)• Black ‐ finished (have found everything reachable from it)


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dfs (G) for (each vertex u V[G]) do

dfsVisit (u)color[u] = gray

color[u] = white π[u] = null

time = 0 for (each vertex u V[G]) doif (color[u] = white) thendfsVisit(u)

d[u] = time = time+1for (each v Adj[u]) doif (color[v] = white) thenπ[v] = udfsVisit(v)

color[u] = blackf[u] = time = time+1


• No source vertex given.• Will explore every edge.

Complexity: Θ(V + E).

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• When one vertex is a descendant of another i th f t th t t t d b th DFSin the forest that was constructed by the DFS.– Parenthesis Theorem

– White‐path Theorem


Parenthesis Theorem• Theorem (Parenthesis theorem):

– For all u, v, exactly one of the following holds:1 d[u] < f [u] < d[v] < f [v] or d[v] < f [v] < d[u] < f [u] and neither of u and v is a1. d[u] < f [u] < d[v] < f [v] or d[v] < f [v] < d[u] < f [u] and neither of u and v is a

descendant of the other.2. d[u] < d[v] < f [v] < f [u] and v is a descendant of u.3. d[v] < d[u] < f [u] < f [v] and u is a descendant of v.

– So d[u] < d[v] < f [u] < f [v] cannot happen.

• Like parentheses:– OK: ( ) [ ] ( [ ] ) [ ( ) ]– Not OK: ( [ ) ] [ ( ] )

• Corollary (Nesting of descendants’ intervals):– v is a proper descendant of u if and only if d[u] < d[v] < f [v] < f [u].


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White‐path Theorem

• Theorem (White‐path theorem):– v is a descendant of u if and only if at time d[u], there is a path u v consisting of only white vertices (Except for u, which was just colored gray)


Classification of Edges• Tree edge:

– In the constructed forest.– Found by exploring (u, v).k d• Back edge:

– (u, v), where u is a descendant of v.• Forward edge:

– (u, v), where v is a descendant of u, but not a tree edge.• Cross edge:

– any other edge.– Can go between vertices in same depth‐first tree or in different depth‐

first trees.

• In an undirected graph, there may be some ambiguity since (u, v) & (v, u) are the same edge.

Cl if b th fi t t b th t t h– Classify by the first type above that matches.

• Theorem:– In DFS of an undirected graph, we get only tree and back edges. No

forward or cross edges.


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Classification of Edges

• Edge (u, v) can be classified by the color of v h th d i fi t l dwhen the edge is first explored:

– WHITE ‐ indicates a tree edge

– GRAY ‐ indicates a back edge

– BLACK indicates a forward or cross edge.• (u, v) is a forward edge if d[u] < d[v]

• (u, v) is a cross edge if d[u] > d[v].


Detection of Cycles

• Lemma:– A directed graph G is acyclic if and only if a DFS of G yields no back edges.

• Proof:– Back edge Cycle

• Suppose there is a back edge (u, v).pp g ( )

• Then v is ancestor of u in the constructed forest.

• Therefore, there is a path v u, so v u → v is a cycle.


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Detection of Cycles

• Cycle back edge.S G t i l C– Suppose G contains cycle C.

– Let v be the first vertex discovered in c, and let (u, v) be the preceding edge in C.

– At time d[v], vertices of C form a white path v u• since v is the first vertex discovered in c.

– By white‐path theorem, u is descendant of v in depth‐first forest.

– Therefore, (u, v) is a back edge.


Topological Sort

• Directed acyclic graph (dag) is a directed graph with no cycles

• Good for modeling a partial order:– a > b and b > c a > c.– May have a and b such that neither a > b nor b > c.

• Topological sort of a dag: a linear ordering of vertices such that if (u, v) E, then u appears somewhere before v.


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Topological Sort

topologicalSort (V, E)

// Assume G is a DAGComplexity: Θ(V + E)

Call dfs(V, E) to compute finishing times f[v] for all v VOutput vertices in order of decreasing finish times


Topological Sort

• Correctness:– Just need to show if (u v) E then f [v] < f [u]Just need to show if (u, v) E, then f [v] < f [u].

• When we explore (u, v), what are the colors of u and v?– u is gray.– v can’t be gray.

• Because then (u, v) is a back edge, but G is a dag.

– If v is white.• By parenthesis theorem d[u] < d[v] < f[v] < f[u]• By parenthesis theorem, d[u] < d[v] < f[v] < f[u].

– If v is black.• Then v is already finished, but u doesn't, therefore, f [v] < f [u].


Data Structures...7/12/2008 6 Graphs • In an undirected graph, G, two vertices v1 and v2 are said...

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